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How many mol of sodium hydroxide are required to make 1.35 L of 2.50 mL solution?

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Nata [24]3 years ago

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3 and a half mL because

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In this case, the undergoing chemical reaction is:

$CuS+\frac{3}{2} O_2\rightarrow CuO+SO_2$

Thus, given the 1.00-kg of 12.5% ore, we can compute the theoretical yield of copper (II) oxide via stoichiometry:

$m_{CuO}^{theoretical}=1.00kgCuS*\frac{1000gCuS}{1kgCuS} *\frac{12.5gCuS}{100gCuS} *\frac{1molCuS}{95.6gCuS} *\frac{1molCuO}{1molCuS} *\frac{79.5gCuO}{1molCuO} \\\\m_{CuO}^{theoretical}=103.95gCuO$

Whereas the third factor accounts for the percent purity of the covellite. Then, given the percent yield, we can compute the actual yield by:

$m_{CuO}=103.95gCuO*0.9\\\\m_{CuO}=93.6gCuO$

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